Singly reinforced beam design
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The document presents an analysis of singly reinforced concrete beams according to IS:456:2000, detailing design steps for determining required dimensions, reinforcement areas, and specific examples for beams under different conditions. It includes calculations for two examples of singular beam designs, highlighting aspects like over-reinforcement, and necessary checks for safety. The content is aimed at civil engineering students and professionals for practical applications in beam design.
![CE8501 Design Of Reinforced Cement Concrete Elements
Unit 1-Introduction
Design of singly reinforced beam
[As per IS456:2000]
Presentation by,
P.Selvakumar.,B.E.,M.E.
Assistant Professor,
Department Of Civil Engineering,
Knowledge Institute Of Technology, Salem.
1](https://image.slidesharecdn.com/singlyreinforcedbeam-design-200813104228/85/Singly-reinforced-beam-design-1-320.jpg)
![Area of steel (Ast)
• If Mu < Mu,lim then it is under reinforced section
4
[Refer IS456 Pg.96]
Ast derived from](https://image.slidesharecdn.com/singlyreinforcedbeam-design-200813104228/85/Singly-reinforced-beam-design-4-320.jpg)
![Area of steel (Ast)
• If Mu = Mu,lim then it is balanced section
5
[Refer IS456 Pg.96]
Ast derived from](https://image.slidesharecdn.com/singlyreinforcedbeam-design-200813104228/85/Singly-reinforced-beam-design-5-320.jpg)
![Step 1 : Determination of required dimensions of beam
[IS 456:2000]
8
Mu,lim = 0.36
𝒙𝒖,𝒎𝒂𝒙
𝒅
[1 – 0.42
𝒙𝒖,𝒎𝒂𝒙
𝒅
] b d2 fck
130*106 = 0.36 * 0.48 [1 – 0.42 ∗ 0.48 ] * 250 * d2 * 20
d = 434.11 mm
d ≈ 460 mm
Assume cover of 40mm, Hence overall depth D = 500mm](https://image.slidesharecdn.com/singlyreinforcedbeam-design-200813104228/85/Singly-reinforced-beam-design-8-320.jpg)
![Step 2 : Classification of beam
[IS 456:2000]
9
For Fe 415 HYSD bars,
Mu,lim = 0.36
𝒙𝒖,𝒎𝒂𝒙
𝒅
[1 – 0.42
𝒙𝒖,𝒎𝒂𝒙
𝒅
] b d2 fck
Mu,lim = 0.36 * 0.48 [1 – 0.42 ∗ 0.48 ] * 250 * 4602 * 20
= 146 x106 N.mm
= 146 kN.m > 130 kN.m [Mu<Mu,lim Under reinforced section]](https://image.slidesharecdn.com/singlyreinforcedbeam-design-200813104228/85/Singly-reinforced-beam-design-9-320.jpg)
![Step 3 : Area of tensile reinforcement (Ast)
[IS 456:2000]
For under reinforced section, Ast derived from Mu
Mu = 0.87 fy Ast d [1 -
𝑨 𝒔𝒕
𝒇𝒚
𝒃 𝒅 𝒇𝒄𝒌
]
130 * 106 = 0.87 * 415 * Ast * 460 [1-
Ast∗415
250∗460∗20
]
= (- 29.97 Ast
2 )+ (166.08 * 103 Ast ) – (130 * 106)
Ast = 943.34 mm2
10](https://image.slidesharecdn.com/singlyreinforcedbeam-design-200813104228/85/Singly-reinforced-beam-design-10-320.jpg)
![Step 6 : Check for Ast[IS 456:2000]
13
[Refer IS456 Pg.46,47]](https://image.slidesharecdn.com/singlyreinforcedbeam-design-200813104228/85/Singly-reinforced-beam-design-13-320.jpg)
![Step 5 : Check for Ast[IS 456:2000]
• Minimum tension reinforcement required
𝐴 𝑠𝑡
𝑏𝑑
=
0.85
𝑓 𝑦
𝐴 𝑠𝑡
250∗460
=
0.85
415
Ast = 235.54 mm2 < 943.34 mm2 Hence safe
14
[Refer IS456 Pg.47]](https://image.slidesharecdn.com/singlyreinforcedbeam-design-200813104228/85/Singly-reinforced-beam-design-14-320.jpg)
![Step 1 : Determination of required dimensions of beam
[IS 456:2000]
18
Mu,lim = 0.36
𝒙𝒖,𝒎𝒂𝒙
𝒅
[1 – 0.42
𝒙𝒖,𝒎𝒂𝒙
𝒅
] b d2 fck
315*106 = 0.36 * 0.48 [1 – 0.42 ∗ 0.48 ] * 230 * d2 * 25
d = 630.14 mm
d ≈ 650mm
Assume cover of 50mm, Hence overall depth D = 700mm](https://image.slidesharecdn.com/singlyreinforcedbeam-design-200813104228/85/Singly-reinforced-beam-design-18-320.jpg)
![Step 2 : Classification of beam
[IS 456:2000]
19
For Fe 415 HYSD bars,
Mu,lim = 0.36
𝒙𝒖,𝒎𝒂𝒙
𝒅
[1 – 0.42
𝒙𝒖,𝒎𝒂𝒙
𝒅
] b d2 fck
Mu,lim = 0.36 * 0.48 [1 – 0.42 ∗ 0.48 ] * 230 * 6502 * 25
= 335.17 x106 N.mm
= 335.17 kN.m > 315 kN.m [Mu<Mu,lim Under reinforced section]](https://image.slidesharecdn.com/singlyreinforcedbeam-design-200813104228/85/Singly-reinforced-beam-design-19-320.jpg)
![Step 3 : Area of tensile reinforcement (Ast)
[IS 456:2000]
For under reinforced section, Ast derived from Mu
Mu = 0.87 fy Ast d [1 -
𝑨 𝒔𝒕
𝒇𝒚
𝒃 𝒅 𝒇𝒄𝒌
]
315 * 106 = 0.87 * 415 * Ast * 650 [1-
Ast∗415
230∗650∗25
]
= (- 26.06 Ast
2 )+ (243.68 * 103 Ast ) – (315 * 106)
Ast = 1549 mm2
Ast = 1550 mm2
20](https://image.slidesharecdn.com/singlyreinforcedbeam-design-200813104228/85/Singly-reinforced-beam-design-20-320.jpg)
![Step 6 : Check for Ast , d, Mu
Mu = 0.87 fy Ast d [1 -
𝑨 𝒔𝒕
𝒇𝒚
𝒃 𝒅 𝒇𝒄𝒌
]
= 0.87 * 415 * 1885 * 640 [1 -
1885 ∗ 415
25 ∗ 230 ∗640
]
Mu = 344.4 x 106 N.mm.
Mu = 344.4 kN.m > 315 kN.m
Hence safe
24](https://image.slidesharecdn.com/singlyreinforcedbeam-design-200813104228/85/Singly-reinforced-beam-design-24-320.jpg)
Singly reinforced beam design
- 1. CE8501 Design Of Reinforced Cement Concrete Elements Unit 1-Introduction Design of singly reinforced beam [As per IS456:2000] Presentation by, P.Selvakumar.,B.E.,M.E. Assistant Professor, Department Of Civil Engineering, Knowledge Institute Of Technology, Salem. 1
- 2. Singly Reinforced Rectangular beam – Analysis 1. Design steps as per IS:456:2000 2. Reinforcement details 3. Example#06 4. Example#07 2
- 3. Design Steps – Singly reinforced concrete beam as per IS:456:2000 1. Determination of required dimensions of beam 2. Classification of beam as Under reinforced or Balanced or Over reinforced section. 3. Finding area of reinforcement in tension zone only (Ast). 4. Providing suitable rebar based on Ast. 5. Determination of depth of NA. 6. Check for Ast. 3
- 4. Area of steel (Ast) • If Mu < Mu,lim then it is under reinforced section 4 [Refer IS456 Pg.96] Ast derived from
- 5. Area of steel (Ast) • If Mu = Mu,lim then it is balanced section 5 [Refer IS456 Pg.96] Ast derived from
- 6. Limiting Moment of resistance • If Mu > Mu,lim then it is over reinforced section • The section must be designed as doubly reinforced section. 6
- 7. Problem#06 Deign of singly reinforced rectangular beam • Design a singly reinforced concrete beam of width 250 mm, subjected to an ultimate moment of 130 kNm. Assume fck =20 MPa and fy = 415 MPa. Given : b = 250mm d = ? M-20 – fck = 20N/mm2 Fe415 – fy = 415N/mm2 Ast = ? Mu= 130 kNm = 130 x 106 N.mm 7 b= 250mm d =? Ast
- 8. Step 1 : Determination of required dimensions of beam [IS 456:2000] 8 Mu,lim = 0.36 𝒙𝒖,𝒎𝒂𝒙 𝒅 [1 – 0.42 𝒙𝒖,𝒎𝒂𝒙 𝒅 ] b d2 fck 130*106 = 0.36 * 0.48 [1 – 0.42 ∗ 0.48 ] * 250 * d2 * 20 d = 434.11 mm d ≈ 460 mm Assume cover of 40mm, Hence overall depth D = 500mm
- 9. Step 2 : Classification of beam [IS 456:2000] 9 For Fe 415 HYSD bars, Mu,lim = 0.36 𝒙𝒖,𝒎𝒂𝒙 𝒅 [1 – 0.42 𝒙𝒖,𝒎𝒂𝒙 𝒅 ] b d2 fck Mu,lim = 0.36 * 0.48 [1 – 0.42 ∗ 0.48 ] * 250 * 4602 * 20 = 146 x106 N.mm = 146 kN.m > 130 kN.m [Mu<Mu,lim Under reinforced section]
- 10. Step 3 : Area of tensile reinforcement (Ast) [IS 456:2000] For under reinforced section, Ast derived from Mu Mu = 0.87 fy Ast d [1 - 𝑨 𝒔𝒕 𝒇𝒚 𝒃 𝒅 𝒇𝒄𝒌 ] 130 * 106 = 0.87 * 415 * Ast * 460 [1- Ast∗415 250∗460∗20 ] = (- 29.97 Ast 2 )+ (166.08 * 103 Ast ) – (130 * 106) Ast = 943.34 mm2 10
- 11. Step 4 : Selecting Rebar size Area for tension zone Ast = 943.3 mm2 Assume 20mm dia bar, No.of bars = 943.3 314.2 = 3 Hence use 3 numbers of 20mm dia bar Area provided, Ast = 942.6 mm2 Hanger bar size Use 8mm dia bar as hanger bar 11 Area of rebar Area = 𝜋 4 (𝑑2) = 50.3 mm2 (8mm ϕ) = 78.5 mm2 (10mm ϕ) = 113.1 mm2 (12mm ϕ) = 201.1 mm2 (16mm ϕ) = 314.2 mm2 (20mm ϕ) = 490.9 mm2 (25mm ϕ) = 804.2 mm2 (32mm ϕ)
- 12. Step 5 : Depth of neutral axis 12 𝑥 𝑢 = 0.87 𝑓𝑦 𝐴𝑠𝑡 0.36 𝑓𝑐𝑘 𝑏 𝑥 𝑢= 0.87 ∗415 ∗ 942.6 0.36 ∗20 ∗250 𝑥 𝑢= 189 mm
- 13. Step 6 : Check for Ast[IS 456:2000] 13 [Refer IS456 Pg.46,47]
- 14. Step 5 : Check for Ast[IS 456:2000] • Minimum tension reinforcement required 𝐴 𝑠𝑡 𝑏𝑑 = 0.85 𝑓 𝑦 𝐴 𝑠𝑡 250∗460 = 0.85 415 Ast = 235.54 mm2 < 943.34 mm2 Hence safe 14 [Refer IS456 Pg.47]
- 15. Step 6 : Reinforcement details of Singly reinforced beam 15 N A Compression Zone Tension Zone d= 460mm Effective cover = 40mm b= 250mm 20mm dia main bar 8mm dia hanger bar Cross section xu= 189 mm D= 500mm
- 16. Assignment#05 • Design a beam to carry a factored moment of 145kNm using grade of M25 and Fe415. (Assume b= 230mm) 16
- 17. Problem#07 Deign of singly reinforced rectangular beam • Design a singly reinforced concrete beam subjected to an ultimate moment of 315 kNm. Assume fck = 25 N/mm2 and fy= 415 N/mm2. In this beam, due to architectural considerations, the width has to be restricted to 230 mm. Given : b = 230mm d = ? M-25 – fck = 25N/mm2 Fe415 – fy = 415N/mm2 Ast = ? Mu= 315 kNm = 315 x 106 N.mm 17 b= 230mm d =? Ast
- 18. Step 1 : Determination of required dimensions of beam [IS 456:2000] 18 Mu,lim = 0.36 𝒙𝒖,𝒎𝒂𝒙 𝒅 [1 – 0.42 𝒙𝒖,𝒎𝒂𝒙 𝒅 ] b d2 fck 315*106 = 0.36 * 0.48 [1 – 0.42 ∗ 0.48 ] * 230 * d2 * 25 d = 630.14 mm d ≈ 650mm Assume cover of 50mm, Hence overall depth D = 700mm
- 19. Step 2 : Classification of beam [IS 456:2000] 19 For Fe 415 HYSD bars, Mu,lim = 0.36 𝒙𝒖,𝒎𝒂𝒙 𝒅 [1 – 0.42 𝒙𝒖,𝒎𝒂𝒙 𝒅 ] b d2 fck Mu,lim = 0.36 * 0.48 [1 – 0.42 ∗ 0.48 ] * 230 * 6502 * 25 = 335.17 x106 N.mm = 335.17 kN.m > 315 kN.m [Mu<Mu,lim Under reinforced section]
- 20. Step 3 : Area of tensile reinforcement (Ast) [IS 456:2000] For under reinforced section, Ast derived from Mu Mu = 0.87 fy Ast d [1 - 𝑨 𝒔𝒕 𝒇𝒚 𝒃 𝒅 𝒇𝒄𝒌 ] 315 * 106 = 0.87 * 415 * Ast * 650 [1- Ast∗415 230∗650∗25 ] = (- 26.06 Ast 2 )+ (243.68 * 103 Ast ) – (315 * 106) Ast = 1549 mm2 Ast = 1550 mm2 20
- 21. Step 4 : Selecting Rebar size Area for tension zone Ast = 1550 mm2 Assume 20mm dia bar, No.of bars = 1550 314.2 = 4.9 Hence use 6 numbers of 20mm dia bar Area provided, Ast = 1885.2 mm2 > 1550 mm2 Hence safe 21 Area of rebar Area = 𝜋 4 (𝑑2) = 50.3 mm2 (8mm ϕ) = 78.5 mm2 (10mm ϕ) = 113.1 mm2 (12mm ϕ) = 201.1 mm2 (16mm ϕ) = 314.2 mm2 (20mm ϕ) = 490.9 mm2 (25mm ϕ) = 804.2 mm2 (32mm ϕ)
- 22. Step 5 : Depth of neutral axis 22 𝑥 𝑢 = 0.87 𝑓𝑦 𝐴𝑠𝑡 0.36 𝑓𝑐𝑘 𝑏 𝑥 𝑢= 0.87 ∗415 ∗1885.2 0.36 ∗25 ∗230 𝑥 𝑢= 328.8 mm
- 23. Step 5 : Reinforcement details of Singly reinforced beam 23 N A d= 640mm Clear cover = 30mm b= 230mm 20mm dia main bar D= 700mm Cross section xu= 328.8 mm Spacing 20 mm • 6 bars cannot be provided at single layer • Hence has to provide at 2 layers • Assume clear cover of 30mm Spacing of 20mm • Effective depth d = 700-30-20-10 d= 640 mm
- 24. Step 6 : Check for Ast , d, Mu Mu = 0.87 fy Ast d [1 - 𝑨 𝒔𝒕 𝒇𝒚 𝒃 𝒅 𝒇𝒄𝒌 ] = 0.87 * 415 * 1885 * 640 [1 - 1885 ∗ 415 25 ∗ 230 ∗640 ] Mu = 344.4 x 106 N.mm. Mu = 344.4 kN.m > 315 kN.m Hence safe 24
- 25. Thank You 25